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Delta function integral

  1. Sep 22, 2007 #1

    nicksauce

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    Evaluate:

    [tex]\int^{\infty}_{-\infty} f(x)\delta(x-x_0)dx[/tex]

    Where

    [tex]f(x)=ln(x+3), x_0=-2[/tex]

    Ordinarily, you would just evaluate [tex]f(x_0)[/tex], so it would be 0, but in this case, since [tex]f(x)[/tex] is [tex]-\infty[/tex] at [tex]x=-3[/tex], does that make a difference?
     
    Last edited: Sep 22, 2007
  2. jcsd
  3. Sep 22, 2007 #2
    But you have x0=-2 and f(x) is certainly not negative infinity there. What is f(-2)?
     
  4. Sep 22, 2007 #3

    nicksauce

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    Right, f(-2) = ln(1) = 0

    I'm just wondering if for the formula:

    [tex]\int_{-\infty}^{\infty}f(x)\delta (x-x_0) = f(x_0) [/tex]

    there need to be certain conditions on f(x), that would make a logarithmic function not suitable? My textbook was very vague in this area.
     
    Last edited: Sep 22, 2007
  5. Sep 22, 2007 #4

    Hurkyl

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    You're missing the + and the - on your limits...


    There are ways to define integration of generalized functions so that this integral makes sense for any (partial) function f that is well-behaved near x0. It might even be possible to make this integral well-defined when f is a generalized function that is well-behaved near x0!
     
  6. Sep 22, 2007 #5

    HallsofIvy

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    No. For any function defined at x0,
    [tex]\int_{-\infty}^\infty f(x)\delta (x)dx= f(x_0)[/tex]
     
  7. Sep 22, 2007 #6

    Hurkyl

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    I think I won't feel comfortable unless you insist on continuity in a neighborhood of x0.
     
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